Chapter 2: Linear Equations in One Variable

Q11Written answer

Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?

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Worked solution

Let the age of Baichung’s father be x.

Then, the age of Baichung’s grandfather = (x+26)

and, the age of Baichung = (x-29). According to the question,

x + (x+26) + (x-29) = 135

⇒ 3x + 26 – 29 = 135

⇒ 3x – 3 = 135

⇒ 3x = 135 + 3

⇒ 3x = 138

⇒ x = 138/3

⇒ x = 46

Age of Baichung’s father = x = 46

Age of Baichung’s grandfather = (x+26) = 46 + 26 = 72

Age of Baichung = (x-29) = 46 – 29 = 17

Q12Written answer

Fifteen years from now, Ravi’s age will be four times his present age. What is Ravi’s present age?

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Worked solution

Let the present age of Ravi be x.

Fifteen years later, Ravi’s age will be x+15 years. According to the question,

x + 15 = 4x

⇒ 4x – x = 15

⇒ 3x = 15

⇒ x = 15/3

⇒ x = 5

Therefore, the present age of Ravi = 5 years.

Q13Written answer

A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get -7/12. What is the number?

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Worked solution

Let the rational be x.

According to the question,

x × (5/2) + 2/3 = -7/12

⇒ 5x/2 + 2/3 = -7/12

⇒ 5x/2 = -7/12 – 2/3

⇒ 5x/2 = (-7- 8)/12

⇒ 5x/2 = -15/12

⇒ 5x/2 = -5/4

⇒ x = (-5/4) × (2/5)

⇒ x = – 10/20

⇒ x = -½

Therefore, the rational number is -½.

Q14Written answer

Lakshmi is a cashier in a bank. She has currency notes of denominations ₹100, ₹50 and ₹10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is ₹4,00,000. How many notes of each denomination does she have?

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Worked solution

Let the numbers of notes of ₹100, ₹50 and ₹10 be 2x, 3x and 5x, respectively.

Value of ₹100 = 2x × 100 = 200x

Value of ₹50 = 3x × 50 = 150x

Value of ₹10 = 5x × 10 = 50x

According to the question,

200x + 150x + 50x = 4,00,000

⇒ 400x = 4,00,000

⇒ x = 400000/400

⇒ x = 1000

Numbers of ₹100 notes = 2x = 2000

Numbers of ₹50 notes = 3x = 3000

Numbers of ₹10 notes = 5x = 5000

Q15Written answer

I have a total of ₹300 in coins of denomination ₹1, ₹2 and ₹5. The number of ₹2 coins is 3 times the number of ₹5 coins. The total number of coins is 160. How many coins of each denomination are with me?

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Worked solution

Let the number of ₹5 coins be x.

Then,

Number ₹2 coins = 3x

And, number of ₹1 coins = (160 – 4x) Now,

Value of ₹5 coins = x × 5 = 5x

Value of ₹2 coins = 3x × 2 = 6x

Value of ₹1 coins = (160 – 4x) × 1 = (160 – 4x)

According to the question,

5x + 6x + (160 – 4x) = 300

⇒ 11x + 160 – 4x = 300

⇒ 7x = 140

⇒ x = 140/7

⇒ x = 20

Number of ₹5 coins = x = 20

Number of ₹2 coins = 3x = 60

Number of ₹1 coins = (160 – 4x) = 160 – 80 = 80

Q16Written answer

The organisers of an essay competition decide that a winner in the competition gets a prize of ₹100 and a participant who does not win gets a prize of ₹25. The total prize money distributed is ₹3,000. Find the number of winners, if the total number of participants is 63.

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Worked solution

Let the number of winners be x.

Then, the number of participants who didn’t win = 63 – x

Total money given to the winner = x × 100 = 100x

Total money given to the participant who didn’t win = 25 × (63-x)

According to the question,

100x + 25 × (63-x) = 3,000

⇒ 100x + 1575 – 25x = 3,000

⇒ 75x = 3,000 – 1575

⇒ 75x = 1425

⇒ x = 1425/75

⇒ x = 19

Therefore, the numbers of winners are 19.

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Exercise 2.2